22-Mec-A4 Design and Manufacture of Machine Elements · Undated paper

Question 4 of 6: Hollow input shaft of a crane gear box — fatigue, slopes and key stresses

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, May 2019 — 16-Mec-A4 Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator. Six questions in two parts: Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine-element analysis). The rubric asks for two questions from each part; all six are solved here. All questions carry equal value (25 %).

Reference texts. S. Kalpakjian and S. Schmid, Manufacturing Engineering and Technology, 7th ed. (Part A: Ch. 16 sheet-metal forming, Ch. 26 grinding); M. Groover, Fundamentals of Modern Manufacturing, 6th ed. (Ch. 20, 25); R. Budynas and K. Nisbett, Shigley's Mechanical Engineering Design, 10th ed. (Ch. 3 stress, Ch. 6 fatigue, Ch. 7 shafts and keys, Ch. 16 brakes); R. Hibbeler, Mechanics of Materials, 10th ed. (Ch. 6, 7, 9 transverse shear and stress transformation).

Question 4: Hollow input shaft of a crane gear box — fatigue, slopes and key stresses (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The shaft is a simply supported hollow tube carrying a pinion at C, with the data below read from page 4 of the paper.

Given data — Q4 hollow input shaft
Quantity	Symbol	Value
Tangential gear force	$F_t$	206 N
Radial gear force	$F_r$	75 N
Transmitted torque	$T$	2.06 N·m
Bearing A to pinion	$a$	66 mm
Pinion to bearing B	$b$	84 mm
Bearing span	$L$	150 mm
Shaft outside diameter	$D$	12 mm
Shaft bore	$d$	6 mm
Pinion pitch diameter	$d_P$	20 mm
Key width (square key), length	$w,\ L_k$	2.4 mm, 25 mm
Ambient air temperature	$T_{\text{amb}}$	$50\ ^\circ\text{C}$

Find. (a) the DE-Goodman fatigue factor of safety at the keyseat, (b) the slope of the deflected shaft at each bearing, checked against rolling-bearing limits, and (c) the shear and bearing stresses carried by the key.

Q4 — hollow input shaft: elevation showing the two gear force components at the pinion C, and the keyed hollow section (12 mm o.d., 6 mm bore, 2.4 mm square key).

Check: the shaft material is not stated on the paper. An open-book fatigue calculation cannot proceed without $S_{ut}$, so the solution adopts AISI 1018 cold-drawn steel, $S_{ut} = 440\ \text{MPa}$, $S_y = 370\ \text{MPa}$, the usual default for a small commercial gearbox shaft, and states it as an assumption exactly as instruction 1 of the paper invites. The method below is independent of that choice; only the numerical factors of safety move. A useful sensitivity: switching to 1035 CD ($S_{ut} = 550\ \text{MPa}$) raises the fatigue factor of safety from 1.54 to 1.83, and 1018 hot-rolled ($S_{ut} = 400\ \text{MPa}$) lowers it to 1.43. The conclusion — adequate but not generous — is unchanged across that range.

Approach. Resolve the gear force into its two planes and combine the bending moments at the pinion, where the keyseat also sits; size the hollow section; build the Marin endurance limit for the stated 50 °C environment; then apply the distortion-energy (von Mises) stresses to the Goodman line, treating the bending as fully reversed and the torque as steady. Slopes come from the standard simply-supported result, and the key is checked in direct shear and in bearing.

Bending moment at the pinion. For a simply supported shaft with a single transverse load $W$ at distance $a$ from A, the moment peaks under the load at $M = Wab/L$. Applying that in each plane, $$M_z = \frac{F_t\,a\,b}{L} = \frac{(206)(66)(84)}{150} = 7614\ \text{N}\cdot\text{mm},\qquad M_y = \frac{F_r\,a\,b}{L} = 2772\ \text{N}\cdot\text{mm}$$ and because the two planes are orthogonal the resultant is their vector sum, $$M = \sqrt{M_z^2 + M_y^2} = \sqrt{7614^2 + 2772^2} = \boxed{8.103\ \text{N}\cdot\text{m}}$$
Hollow section properties. With $D = 12\ \text{mm}$ and $d = 6\ \text{mm}$, $$I = \frac{\pi\left(D^4 - d^4\right)}{64} = \frac{\pi\left(12^4-6^4\right)}{64} = 954.26\ \text{mm}^4,\qquad J = 2I = 1908.5\ \text{mm}^4$$ so the section moduli are $Z = I/(D/2) = 159.04\ \text{mm}^3$ in bending and $Z_p = J/(D/2) = 318.09\ \text{mm}^3$ in torsion. Removing the bore costs only 6 % of the solid $I$ while saving 25 % of the mass — the reason the shaft is hollow at all.
Nominal stresses at the keyseat. The shaft rotates, so a fixed point on its surface sees the bending moment as a fully reversed stress, while the transmitted torque is steady: $$\sigma_a = \frac{M}{Z} = \frac{8103}{159.04} = 50.95\ \text{MPa},\qquad \tau_m = \frac{T}{Z_p} = \frac{2060}{318.09} = 6.476\ \text{MPa}$$ with $\sigma_m = 0$ and $\tau_a = 0$.
Endurance limit by the Marin factors. Starting from $S'_e = 0.5S_{ut} = 220\ \text{MPa}$: surface (machined/cold-drawn) $k_a = 4.51\,S_{ut}^{-0.265} = 0.899$; size $k_b = 1.24\,D^{-0.107} = 0.951$; load $k_c = 1$ because the distortion-energy combination is handled explicitly; temperature at $50\ ^\circ\text{C}$ ($122\ ^\circ\text{F}$) $k_d = 1.012$ — slightly above unity, as steels gain a little fatigue strength just above room temperature; reliability $k_e = 1$ at 50 %. Hence $$S_e = k_ak_bk_ck_dk_e S'_e = (0.899)(0.951)(1)(1.012)(1)(220) = \boxed{190.3\ \text{MPa}}$$
Stress concentration at the keyseat. The critical section is the keyseat itself, not the bearing journals, because it combines the maximum moment with a sharp notch. For an end-milled keyseat, Shigley Table 7-1 gives $K_t = 2.14$ in bending and $K_{ts} = 3.0$ in torsion; taking $K_f = K_t$ and $K_{fs} = K_{ts}$ (full notch sensitivity) is the conservative and customary choice for a keyseat.
Distortion-energy (von Mises) alternating and midrange stresses. With reversed bending only on the alternating side and steady torque only on the midrange side, $$\sigma'_a = K_f\sigma_a = (2.14)(50.95) = 109.03\ \text{MPa},\qquad \sigma'_m = \sqrt{3}\,K_{fs}\tau_m = \sqrt{3}\,(3.0)(6.476) = 33.65\ \text{MPa}$$
Apply the Goodman line. The modified-Goodman criterion combines the two through $$\frac{1}{n} = \frac{\sigma'_a}{S_e} + \frac{\sigma'_m}{S_{ut}} = \frac{109.03}{190.3} + \frac{33.65}{440} = 0.5730 + 0.0765 = 0.6495$$ $$\boxed{n = 1.54}$$ The shaft has infinite life with about a 50 % margin. Note where that margin is spent: the reversed bending accounts for 88 % of the damage sum and the steady torque for only 12 %, so this is a bending-driven design and any change to the gear separating loads or the span matters far more than a change in torque.
Check first-cycle yielding. The peak von Mises stress in a single load cycle is $$\sigma'_{\max} = \sqrt{(K_f\sigma_a)^2 + 3(K_{fs}\tau_m)^2} = \sqrt{109.03^2 + 3(19.43)^2} = 114.1\ \text{MPa}$$ giving $n_y = S_y/\sigma'_{\max} = 370/114.1 = \boxed{3.24}$. Yielding is not the governing mode; fatigue is, which is the expected result for a rotating shaft and confirms that the Goodman number is the one to report.
(b) Slopes at the bearings. Both transverse forces act at the same station, so their resultant $F = \sqrt{206^2+75^2} = 219.2\ \text{N}$ can be used once and the answer is the magnitude of the vector slope. For a simply supported beam with a load at $a$ from A, $$\theta_A = \frac{F\,b\left(L^2-b^2\right)}{6LEI},\qquad \theta_B = \frac{F\,a\left(L^2-a^2\right)}{6LEI}$$ With $E = 207\ \text{GPa}$ and $I = 954.26\ \text{mm}^4$, $$\theta_A = \frac{(219.2)(84)\left(150^2-84^2\right)}{6(150)(207000)(954.26)} = \boxed{1.600\times10^{-3}\ \text{rad}}\qquad \theta_B = \boxed{1.477\times10^{-3}\ \text{rad}}$$ that is, 1.600 and 1.477 milliradians, or 0.092° and 0.085°. The larger slope is at A, the bearing on the longer side of the span, as it must be.
Interpret the slopes. Shigley Table 7-2 gives allowable misalignment of roughly 0.0005–0.0012 rad for cylindrical and tapered roller bearings and 0.001–0.003 rad for deep-groove ball bearings. Both computed slopes exceed the roller-bearing band and sit in the upper half of the ball-bearing band. The shaft is therefore acceptable on deep-groove ball bearings but should not be run in tapered rollers without either self-aligning bearings or a stiffer shaft; since $\theta \propto 1/I$, increasing the outside diameter from 12 to 14 mm at the same bore would cut both slopes by about 46 % and bring them inside the roller limit.
(c) Key stresses. The key transmits the torque at the shaft surface, so the force it carries is $$F_{\text{key}} = \frac{T}{D/2} = \frac{2060}{6} = 343.3\ \text{N}$$ A square key has height $h = w = 2.4\ \text{mm}$. It fails either by shearing across its width or by crushing on the half of its height that bears against the keyseat wall: $$\tau = \frac{F_{\text{key}}}{w L_k} = \frac{343.3}{(2.4)(25)} = \boxed{5.72\ \text{MPa}},\qquad \sigma_{\text{brg}} = \frac{F_{\text{key}}}{(h/2)L_k} = \frac{343.3}{(1.2)(25)} = \boxed{11.44\ \text{MPa}}$$
Assess the key. Against the same 1018 CD properties, the shear factor of safety is $n = 0.577S_y/\tau = 0.577(370)/5.72 = 37.3$ and the bearing factor is $n = S_y/\sigma_{\text{brg}} = 32.3$. The key is more than twenty times stronger than the shaft it drives — which is the correct design outcome only if one accepts that the key here is sized by handling and fit rather than by load. If the key were intended as a deliberate mechanical fuse it would need to be far smaller, and the more likely reading is that a 2.4 mm square key is simply the smallest standard stock size for a 12 mm shaft.

Q4 — final results
Quantity	Symbol	Result
Resultant bending moment at the pinion	$M$	8.103 N·m
Endurance limit (Marin, $50\ ^\circ\text{C}$)	$S_e$	190.3 MPa
von Mises alternating / midrange stress	$\sigma'_a$ / $\sigma'_m$	109.03 / 33.65 MPa
(a) DE-Goodman factor of safety	$n$	1.54
First-cycle yield factor	$n_y$	3.24
(b) Slope at bearing A	$\theta_A$	1.600 mrad (0.092°)
(b) Slope at bearing B	$\theta_B$	1.477 mrad (0.085°)
(c) Key shear stress	$\tau$	5.72 MPa ($n = 37.3$)
(c) Key bearing stress	$\sigma_{\text{brg}}$	11.44 MPa ($n = 32.3$)
Diameter needed for $n = 2.0$ (same bore ratio)	$D$	13.13 mm

Topic. Fatigue design of a rotating shaft by the DE-Goodman criterion, shaft slope at rolling bearings, and key sizing.

Key relations. $M = Wab/L$; $I = \pi(D^4-d^4)/64$; Marin, $S_e = k_ak_bk_ck_dk_eS'_e$ with $S'_e = 0.5S_{ut}$; DE-Goodman, $1/n = \sigma'_a/S_e + \sigma'_m/S_{ut}$ where $\sigma'_a = K_f\sigma_a$ and $\sigma'_m = \sqrt3\,K_{fs}\tau_m$ for reversed bending with steady torque; $\theta_A = Fb(L^2-b^2)/6LEI$; key $\tau = F/wL_k$, $\sigma_{\text{brg}} = 2F/hL_k$ with $F = 2T/D$.

Why this works. A rotating shaft turns a stationary bending moment into a fully reversed surface stress, while a steady drive torque produces a purely midrange shear. Those two live on different axes of the Goodman diagram, which is why they cannot simply be added: the distortion-energy combination collapses each axis to a single equivalent stress first, and only then does the Goodman line interpolate between infinite life at $S_e$ and static fracture at $S_{ut}$. The keyseat governs because it is where the largest moment coincides with the sharpest notch, and the slope check is a separate serviceability limit — a shaft can be perfectly safe in fatigue and still destroy its bearings through misalignment.

Common pitfall. Placing the torque on the alternating axis (it is steady, so it is a midrange stress), and forgetting the $\sqrt3$ when converting a shear stress into a von Mises equivalent. A third trap is computing the bending moment at the bearing or at mid-span; for a simply supported shaft with a single gear it peaks under the gear, which is also where the keyseat is.

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th ed., §6-9 (Marin factors), §6-12 (Goodman), §7-1 to 7-3 (shaft design, Table 7-1 keyseat factors, Table 7-2 slope limits) and §7-7 (keys).

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